Let $H = \{ \sigma\in S_5\mid\sigma(5)=5\}$. Let $K$ be a subgroup of $S_5$. Prove that $HK=S_5$ iff 5 divides $|K|$.

Question

Let $H = \{ \sigma\in S_5\mid\sigma(5)=5\}$. Let $K$ be a subgroup of $S_5$. Prove that $HK=S_5$ iff 5 divides $|K|$.

Attempt: I found $|H|=24$. I know that $|HK|=\frac{|H||K|}{|H \cap K|}$. Suppose $HK = S_5$. Then $|HK|=120$, but I don't know that $H \cap K=\{e\}$ or not. Also how do I show the converse?

Any help is appreciated.

Answer 1

If $5\mid |K|$, then $|K|=5m$. We have $|HK||H\cap K|=120m$. But $|H\cap K|\mid 24$ and $|H\cap K|\mid 5m$, so $$|H\cap K|\mid \gcd(24,5m)=\gcd(24,m)\mid m.$$ Then $|HK|$ is a multiple of $120$ and since $HK\subseteq S_5$ we get $HK=S_5$.

Answer 2

Assume that $5$ divides the order of $K$, subgroup of $S_5$. Then $K$ contains an element of order $5$, which must be a cycle $c$ of this length. Then $5$, $c(5)$, $c^2(5)$, $c^3(5)$, $c^4(5)$ are different elements, so they are in some order $1,2,3,4,5$.

Now consider an element $s$ in $S_5$. (We want to show it is in the set $HK$.) Look at $s(5)$ and pick the power $k$ so that $s(5)=c^k(5)$. Then $sc^{-k}(5)=5$, so $h:=sc^{-k}$ is in $H$. This gives $$ s = hs^k\in HK\ . $$

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Conversely, assume that $HK=S_5$. By definition, $HK$ is the set of all products $hk$ where $h$ is running in $H$ and $k$ in $K$. In our case $HK=KH$ is even a group, the group generated by $H,K$, it has $5!=120$ elements, so from $$ 120=|HK|=\frac{|H|\;|K|}{|H\cap K|} $$ we obtain that $|H|\;|K|=120\;|H\cap K|$. Five divides the RHS, so it also divides the LHS, which is $|H|\;|K|=24|K|$. So five divides $|K|$.